JEE Main 2019 (08 April Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Differential Equations: Given that the slope of the tangent to a curve $y = y (x)$ at any point $(x, y)$ is $2 y / x^{2}$ . If the curve passes through the centre of the circle $x^{2} + y^{2} - 2 x - 2 y = 0$ , then its equation is :

Q: Wait, why do we reject the negative root here or why is there an absolute value in the logarithm?

Great observation! The integral of $1/y$ is $\ln|y|$ because the domain of the natural logarithm function is only positive numbers. When we solve differential equations, we use the absolute value to ensure the expression is defined for all valid $y$. In this specific problem, the curve passes through $(1, 1)$, which is in the first quadrant, so we can safely work with positive values.

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the slope of the tangent as the derivative

d y / d x

. The problem gives you a direct relationship between the slope and the coordinates

(x, y)

. The circle equation is not just a circle; it is a geometric constraint that provides a specific point

(h, k)

through which your curve must pass.

Hint 2

Hint 2: The Arsenal. You are dealing with a first-order differential equation. The most effective tool here is the method of separation of variables. Group all terms involving

y

on one side and all terms involving

x

on the other, then integrate both sides.

Hint 3

Hint 3: The Execution. After separating the variables, you will have

\int \frac{1}{y} d y = \int \frac{2}{x ^{2}} d x

. Integrate these to get

ln ∣ y ∣ = - \frac{2}{x} + C

. Now, use the center of the circle

(1, 1)

to find the value of the constant

C

Visualized Solution

Setting up the Differential Equation

The slope of the tangent to a curve $y = y (x)$ at any point $(x, y)$ is given by $\frac{d y}{d x}$ .
We are given that this slope is $\frac{2 y}{x ^{2}}$ .
Therefore, our differential equation is: $\frac{d y}{d x} = \frac{2 y}{x ^{2}}$ .

Variable Separation Method

To solve $\frac{d y}{d x} = \frac{2 y}{x ^{2}}$ , we use the method of separation of variables.
We need to group all $y$ terms with $d y$ and all $x$ terms with $d x$ .
Rearranging gives: $\frac{1}{y} d y = \frac{2}{x ^{2}} d x$ .

Setting up the Integrals

Apply integration to both sides of the separated equation.
$\int \frac{1}{y} d y = \int \frac{2}{x ^{2}} d x$
We can rewrite the right side using negative exponents: $\int \frac{1}{y} d y = 2 \int x^{- 2} d x$ .

Finding the General Solution

The integral of $\frac{1}{y}$ with respect to $y$ is $ln ∣ y ∣$ .
The integral of $x^{- 2}$ is $\frac{x ^{- 1}}{- 1} = - \frac{1}{x}$ .
Adding the constant of integration $C$ , we get the general solution: $ln ∣ y ∣ = - \frac{2}{x} + C$ .

Finding the Circle's Center

The problem states the curve passes through the center of the circle: $x^{2} + y^{2} - 2 x - 2 y = 0$ .
To find the center, we complete the square for both $x$ and $y$ terms.
Group the terms: $(x^{2} - 2 x) + (y^{2} - 2 y) = 0$ .

Coordinates of the Center

Add $1$ to both groups to complete the squares: $(x^{2} - 2 x + 1) + (y^{2} - 2 y + 1) = 1 + 1$ .
This simplifies to the standard circle equation: $(x - 1)^{2} + (y - 1)^{2} = 2$ .
Comparing with $(x - h)^{2} + (y - k)^{2} = r^{2}$ , the center $(h, k)$ is $(1, 1)$ .

Substituting the Center

We know the curve passes through the center $(1, 1)$ .
Substitute $x = 1$ and $y = 1$ into our general solution: $ln ∣ y ∣ = - \frac{2}{x} + C$ .
This gives: $ln ∣1∣ = - \frac{2}{1} + C$ .

Solving for $C$

Evaluate the terms: $ln (1) = 0$ .
The equation becomes: $0 = - 2 + C$ .
Solving for $C$ , we get: $C = 2$ .

Constructing the Final Curve

Substitute $C = 2$ back into the general solution: $ln ∣ y ∣ = - \frac{2}{x} + 2$ .
To match the given options, let's simplify this expression.
Take the common denominator on the right side: $ln ∣ y ∣ = \frac{- 2 + 2 x}{x}$ .

The Final Answer

We have $ln ∣ y ∣ = \frac{2 ( x - 1 )}{x}$ .
Multiply both sides by $x$ : $x ln ∣ y ∣ = 2 (x - 1)$ .
In mathematics, $ln ∣ y ∣$ is the same as $lo g_{e} ∣ y ∣$ .
Therefore, the final equation is $x lo g_{e} ∣ y ∣ = 2 (x - 1)$ , which matches Option 1.

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The Sigma Insight: Variable Separable Method

Solution Diagram

Analyzing the Setup

Imagine you are standing on a two-dimensional coordinate plane. Before you lies a curve,

y = y (x)

, where the slope of the tangent line at any point

(x, y)

is defined by the differential equation:

\frac{d y}{d x} = \frac{2 y}{x ^{2}}

This equation represents the fundamental growth property of the curve, dictating its behavior at every coordinate.

The Art of Separation

To solve this, we employ the method of separation of variables. We organize the terms to isolate

y

on one side and

x

on the other:

\frac{1}{y} d y = \frac{2}{x ^{2}} d x

Next, we apply the integral operator to both sides of the equation:

\int \frac{1}{y} d y = \int 2 x^{- 2} d x

Performing the integration, the left side yields the natural logarithm, while the right side follows the power rule:

ln ∣ y ∣ = - \frac{2}{x} + C

Here,

C

represents the constant of integration, defining an infinite family of curves that share the same slope property.

The Geometric Anchor

We must identify the specific curve that passes through the center of the circle defined by

x^{2} + y^{2} - 2 x - 2 y = 0

. We find the center by completing the square:

(x^{2} - 2 x + 1) + (y^{2} - 2 y + 1) = 1 + 1

(x - 1)^{2} + (y - 1)^{2} = 2

The center of this circle is clearly at the point

(1, 1)

. This point serves as the anchor to determine our constant

C

Substituting

x = 1

and

y = 1

into our general solution:

ln ∣1∣ = - \frac{2}{1} + C

Since

ln (1) = 0

, the equation simplifies to

0 = - 2 + C

, which reveals that $C = 2$ .

The Final Synthesis

With the value of

C

determined, our specific equation becomes:

ln ∣ y ∣ = - \frac{2}{x} + 2

We simplify the right side by finding a common denominator:

ln ∣ y ∣ = \frac{- 2 + 2 x}{x} = \frac{2 ( x - 1 )}{x}

Multiplying both sides by

x

, we arrive at the final equation of the curve:

$x ln ∣ y ∣ = 2 (x - 1)$

Common Traps & FAQs

Wait, why do we reject the negative root here or why is there an absolute value in the logarithm?

Great observation! The integral of

1/ y

ln ∣ y ∣

because the domain of the natural logarithm function is only positive numbers. When we solve differential equations, we use the absolute value to ensure the expression is defined for all valid

y

. In this specific problem, the curve passes through

(1, 1)

, which is in the first quadrant, so we can safely work with positive values.

Why can't we just use energy conservation or some other physics principle?

That is a common instinct in physics! However, this is a pure mathematical problem about the geometry of a curve. We are given the slope of the tangent, which is the definition of the derivative. We must use the tools of calculus—specifically differential equations—to find the function

y (x)

that satisfies this slope condition.

How do I know for sure that the center of the circle is

(1, 1)

Excellent question. To find the center of a circle from the general form

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

, we complete the square. For

x^{2} - 2 x

, we add

1

to make it

(x - 1)^{2}

. We do the same for

y^{2} - 2 y

. This shifts the origin to

(1, 1)

, which is the center. Always remember to balance the equation by adding the same values to the right side!

Animated Solution for Mathematics - Differential Equations: Given that the slope of the tangent to a curve $y = y (x)$ at any point $(x, y)$ is $2 y / x^{2}$ . If the curve passes through the centre of the circle $x^{2} + y^{2} - 2 x - 2 y = 0$ , then its equation is :

Select Answer:

Visualized Solution

Setting up the Differential Equation

Variable Separation Method

Setting up the Integrals

Finding the General Solution

Finding the Circle's Center

Coordinates of the Center

Substituting the Center

Solving for $C$

Constructing the Final Curve

The Final Answer

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Geometric Anchor

The Final Synthesis

Similar Questions

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Select Answer:

Visualized Solution

Setting up the Differential Equation

Variable Separation Method

Setting up the Integrals

Finding the General Solution

Finding the Circle's Center

Coordinates of the Center

Substituting the Center

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C

Constructing the Final Curve

The Final Answer

The Sigma Insight: Variable Separable Method

Analyzing the Setup

The Art of Separation

The Geometric Anchor

The Final Synthesis

Similar Questions

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Solving for $C$

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